When the 5 -kg box reaches point A it has a speed vA=10 m / s . Determine the normal force the b

step 1 Solving Part A of this problem, so here I can write A is equal to FX by M according to the quest

When the 5 -kg box reaches point A it has a speed vA=10 m /...

When the 5 -kg box reaches point $A$ it has a speed $v_{A}=10 \mathrm{m} / \mathrm{s} .$ Determine the normal force the box exerts on the surface when it reaches point $B$. Neglect friction and the size of the box.

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Key Concepts

Newton's Second Law

This fundamental law relates the net force acting on an object to its mass and acceleration (F = ma). It is applied to determine the correct force balance, incorporating both gravitational and centripetal contributions, to compute the normal force exerted by the box on the surface.

Normal Force

This is the perpendicular contact force exerted by a surface on an object. In dynamics, especially on curved surfaces, the normal force is influenced not only by the gravitational force but also by the requirement to provide the necessary centripetal force for curved motion.

Conservation of Energy

This principle states that in the absence of non-conservative forces like friction, the total mechanical energy of a system (kinetic plus potential energy) remains constant. In the context of the problem, energy conservation is used to relate the speeds and heights of the box at different points along its path.

Centripetal Acceleration

When an object travels along a curved path, it must experience an inward acceleration (centripetal acceleration) given by the square of its speed divided by the radius of curvature. This concept is essential for analyzing the additional forces required to change the direction of the object’s velocity.

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